Question

Two batteries of emf $4V$ and $8V$ with internal resistance $1\Omega$ and $2\Omega$ respectively are connected to an external resistance $R=9\Omega$ as shown in figure. The current in circuit and the potential difference between $P$ and $Q$ respectively will be

• A. $\frac{1}{9}A,9V$
• B. $\frac{1}{2}A,12V$
• C. $\frac{1}{3}A,3V$
• D. $\frac{1}{6}A,4V$

Answer 1

Answer: (C)

$\frac{1}{3}A,3V$

Since the batteries are connected in reverse polarities, the net potential applied to the circuit $=8V-4V=4V$

The net resistance in the circuit $=R+r_1+r_2$ $=9\Omega +1\Omega +2\Omega =12\Omega$

Hence, the current in the circuit $=\frac{4V}{12\Omega }$ $=\frac{1}{3}A$

Potential difference across P and Q $=IR$ $=\frac{1}{3}A\times 9\Omega$ $=3V$